Re yesterday’s puzzle, you’ll find answers in the comments. (We are blessed with some very smart commenters here at The Big Questions!!)
Commenter Roger Schlafly pointed this Wikipedia article where I was surprised and delighted to see a reference to a paper co-written by my old friend Dave Rusin. I did not remember that Dave had anything to do with this problem, but in retrospect I bet I knew this at one time.
I managed to dig out some notes I jotted down on this subject many many years ago. I have not doublechecked these results, and I can’t completely vouch for the careful accuracy of my younger self, so take these for what they’re worth. But here’s what I once claimed to have proved:
The reason there is exactly one pair of nonstandard six-sided dice is that six is the product of two distinct primes. For the same reason, there is exactly one pair of nonstandard n-sided dice when n is 10, or 15, or 21, or …. For any product of three distinct primes, there are at most 40 nonstandard pairs.
I also found (in what appears to be my handwriting) this chart, which I reproduce with the same caveats:
| Number of sides | Number of nonstandard pairs |
| 1 | 0 |
| 2 | 0 |
| 3 | 0 |
| 4 | 1 |
| 5 | 0 |
| 6 | 1 |
| 7 | 0 |
| 8 | 3 |
| 9 | 1 |
| 10 | 1 |
| 11 | 0 |
| 12 | at most 13 |
| 13 | 0 |
| 14 | 1 |
| 15 | 1 |
| 16 | 9 |
| 17 | 0 |
| 18 | at most 13 |
| 19 | 0 |
| 20 | at most 13 |
| 21 | 1 |
| 22 | 1 |
| 23 | 0 |
| 24 | at most 94 |
| 25 | 1 |
| 26 | 1 |
| 27 | 3 |
| 28 | at most 13 |
| 29 | 0 |
| 30 | at most 40 |
| 31 | 0 |
| 32 | 25 |
Corrections welcome!
Challenge : proof or counterexample : Let a(n)=((2*n-1)*a(n-1)+3*(n-1)*a(n-2))/n, a(0)=a(1)=1. Then if n=p^k, there are exactly (a(k)-1)/2 nonstandard pairs of dice.
For example, I claim that for n=1024, there are exactly 4476 nonstandard pairs of dice that give the same probabilities for their sums as a “standard” pair of 1024-sided dice.
Not that I can quite imagine a “standard pair of 1024-sided dice”…
Mike H: I’m sure you meant to assume p is prime!
But I got a different formula. For n=1024 (so that p=10), I get 8953, not 4476. To avoid spoilers, I won’t state my general formula.
yes, p should be prime! But aren’t you counting {A,B} and {B,A} as two different pairs of dice?
I counted {A,B} and {B,A} as the same pair of dice, and didn’t count the standard pair. So I think we have the same formula actually…
Mike H: Ah. It seems we do agree.
You can also buy a pair for yourself!
http://www.grand-illusions.com/acatalog/Sicherman_Dice.html
@Super – that’s super :-)
… so – anyone for a game of backgammon or monopoly?
For people who want a partial spoiler, see http://j.mp/oPyzoP and http://j.mp/pfjW3V
I found 7 pairs for the dodecahedron, if that means anything to anybody… Besides my family… who haven’t seen me in days… Maybe I’ll hold off on the icosahedron…
Tom Temple: This is fabulous. Take a day or two with your family if you must, though…..
I got 12:7, 18:7, 20:7, 24:32, 28:7, 30:12.
I suspect it’s not a coincidence that 12, 18, 20, 28 each have 7 seeing how each also has prime factors p_1^2*p_2.
Following through a bit more, 42 has 12 pair, just like 30. This all seems to suggest that the number is exactly determined by the multiplicities of its factorization. Any mathematicians want to take up this conjecture. Maybe come up with a formula even?